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1630. Arithmetic Subarrays

A sequence of numbers is called arithmetic if it consists of at least two elements, and the difference between every two consecutive elements is the same. More formally, a sequence s is arithmetic if and only if s[i+1] - s[i] == s[1] - s[0] for all valid i.

For example, these are arithmetic sequences:

1, 3, 5, 7, 9 7, 7, 7, 7 3, -1, -5, -9 

The following sequence is not arithmetic:

1, 1, 2, 5, 7 

You are given an array of n integers, nums, and two arrays of m integers each, l and r, representing the m range queries, where the ith query is the range [l[i], r[i]]. All the arrays are 0-indexed.

Return a list ofbooleanelementsanswer, whereanswer[i]istrueif the subarraynums[l[i]], nums[l[i]+1], ... , nums[r[i]]can be rearranged to form an arithmetic sequence, andfalseotherwise.

Example 1:

Input: nums = [4,6,5,9,3,7], l = [0,0,2], r = [2,3,5] Output: [true,false,true] Explanation: In the 0th query, the subarray is [4,6,5]. This can be rearranged as [6,5,4], which is an arithmetic sequence. In the 1st query, the subarray is [4,6,5,9]. This cannot be rearranged as an arithmetic sequence. In the 2nd query, the subarray is [5,9,3,7]. This can be rearranged as [3,5,7,9], which is an arithmetic sequence. 

Example 2:

Input: nums = [-12,-9,-3,-12,-6,15,20,-25,-20,-15,-10], l = [0,1,6,4,8,7], r = [4,4,9,7,9,10] Output: [false,true,false,false,true,true] 

Constraints:

  • n == nums.length
  • m == l.length
  • m == r.length
  • 2 <= n <= 500
  • 1 <= m <= 500
  • 0 <= l[i] < r[i] < n
  • -105 <= nums[i] <= 105

Solutions (Ruby)

1. Sort

# @param {Integer[]} nums# @param {Integer[]} l# @param {Integer[]} r# @return {Boolean[]}defcheck_arithmetic_subarrays(nums,l,r)ret=[false] * l.size(0...l.size).eachdo |i| sub=nums[l[i]..r[i]].sortret[i]=sub.size > 1 && (2...sub.size).all?{ |j| sub[j] - sub[j - 1] == sub[1] - sub[0]}endretend

Solutions (Rust)

1. Sort

implSolution{pubfncheck_arithmetic_subarrays(nums:Vec<i32>,l:Vec<i32>,r:Vec<i32>) -> Vec<bool>{letmut ret = vec![false; l.len()];for i in0..l.len(){letmut sub = nums[l[i]asusize..=r[i]asusize].to_vec(); sub.sort_unstable(); ret[i] = sub.len() > 1 && (2..sub.len()).all(|j| sub[j] - sub[j - 1] == sub[1] - sub[0]);} ret }}
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